The quadratic equation of one variable (3x + m) ^ 2 = n (x-3) ^ 2 is reduced to a general form The univariate quadratic equation (3x + m) ^ 2 = n (x-3) ^ 2 is reduced to a general form, which is a univariate quadratic equation about X, and Mn is a constant

The quadratic equation of one variable (3x + m) ^ 2 = n (x-3) ^ 2 is reduced to a general form The univariate quadratic equation (3x + m) ^ 2 = n (x-3) ^ 2 is reduced to a general form, which is a univariate quadratic equation about X, and Mn is a constant

9x²-6mx+m²=nx²-6nx+9n
So (N-9) x & # 178; + (6m-6n) x + (9n-m & # 178;) = 0

General form of quadratic equation of one variable
I know that the general form is ax + BX + C = 0 (a is not equal to 0)
What is the general form of 5x = 3x-1
Is it 5x-3x + 1 = 0 or 3x-1-5x = 0
Why?
Is it possible for quadratic equation of one variable to have no solution?
If the constant term is 0, can we use the cross phase to solve it?
It is mainly about the general form of 5x = 3x-1
5x - 3x + 1 = 0 - 5x + 3x - 1 = 0
Which is right? Why?

The general form is a habit
The general form of 5x = 3x-1 is 3x-5x-1 = 0
If the constant term is 0, can we use the cross phase to solve it?
Don't you need to extract x directly?
For example, if 3x-5x = 0, extract x (3x-5) = 0
X = O or x = five thirds
To tell you the truth, I have a headache when I look at mathematics (although I'm good at mathematics). It's still a simple knowledge in junior high school, and I can get into trouble when I go to senior high school~

How to change this differential equation into an integral equation?
dy/dx=I*r*(1-N/K)
Where I, R and K are constants
I'm a rookie. Please give me some advice
Wrong, it should be dn / DX = I * r * (1-N / k)

dN/(1-N/K)=lrdx
KdN/(K-N)=lrdx
-Kd(K-N)/(K-N)=lrdx
Integral on both sides at the same time
-Kln(K-N)=lrx+lnC
ln[((K-N)^(-K))/C]=lrx
(K-N)^(-K)=Ce^(lrx)
K-N=Ce^(-lrx/K)
N=K-Ce^(-lrx/K)

How to reduce the elliptic equation of grade two to the standard equation x Λ 2 / 25 + 4Y Λ 2 / 75 = 1

Is Λ square? If so, the equation is as follows:
X ∧ 2 / 25 + y ∧ 2 / 75 / 4 = 1, that is, X ∧ 2 / 25 + y ∧ 2 / 18.75 = 1

Y = 1 / 2x-2 reduced to general equation

Double two on both sides
2y=x-4
x-2y-4=0

A simple equation can be solved directly according to the quantity relation. For a complex equation, the equation with unknowns is regarded as () and reduced to（

A simple equation can be solved directly according to the quantity relationship. For a complex equation, the equation with unknowns is regarded as (equation) and simplified into (simple equation)

I worked out a problem, but I don't know if it's right Please help to see ~ give me an accurate answer! Y = - 3 (x + 3) (x + 9) is reduced to the form of a (x + - b) square + - C

y=-3(x²+12x+27)
=-3[(x²+12x+36)-36+27]
=-3[(x+6)²-9]
=-3(x+6)²+27

How to simplify √ (2 √ 3 + 4) to √ 3 + 1
The root sign is twice the root sign, and the root sign 3 + 4 is the root sign 3 + 1 at the end of the square root. How is this simplified

2√3+4
=3+2√3+1
=(√3)²+2*√3*1+1²
=(√3+1)²
The next formula is √ 3 + 1

Reduced to a * 10 ^ n: 7000 * (- 40000) ^ 2

7000*(-40000)^2
=7000*40000^2
=7*10^3*(4*10^4)^2
=7*10^3*16*10^8
=112*10^11
=1.12*10^13

Simplify a ^ 2 + B ^ 2 + C ^ 2-ab-bc-ac

a²+b²+c²-ab-bc-ac
=(2a²+2b²+2c²-2ab-2bc-2ac)/2
=[(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ac+a²)]/2
=[(a-b)²+(b-c)²+(c-a)²]/2